2 Marks Questions - Maths STD 9 Questions - Vidyadip

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38 questions · timed · auto-graded

Question 12 Marks

Factorise the following:
$8\text{p}^3+\frac{12}{5}\text{p}^2+\frac{6}{25}\text{p}+\frac{1}{125}$

Answer

$8\text{p}^3+\frac{12}{5}\text{p}^2+\frac{6}{25}\text{p}+\frac{1}{125}$
$=(2\text{p})^3+3\times(2\text{p})^2\times\frac{1}{5}+3\times(2\text{p})\times\Big(\frac{1}{5}\Big)^2+\Big(\frac{1}{5}\Big)^3$
$=\Big(2\text{p}+\frac{1}{5}\Big)^3$
[Using identity, (a - b)³ = a³ - b³ + 3a(-b)(a - b)]
$=\Big(2\text{p}+\frac{1}{5}\Big)\Big(2\text{p}+\frac{1}{5}\Big)\Big(2\text{p}+\frac{1}{5}\Big)$
$=\Big(2\text{p}+\frac{1}{5}\Big)^3$

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Question 22 Marks

Expand the following:
(3a - 2b)³

Answer

(3a - 2b)³ = (3a)³ + (-2b)³ + 3(3a)(-2b)(3a - 2b)
[Using identity, (a - b)³ = a³ - b³ + 3a(-b)(a - b)]
= 27a³ - 8b³ - 18ab(3a - 2b)
= 27a³ - 8b³ - 54a²b + 36ab²
= 27a³ - 54a²b + 36ab² - 8b³

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Question 32 Marks

Factorise the following:
16x² + 4y² + 9z² - 16xy - 12yz + 24xz

Answer

16x² + 4y² + 9z² - 16xy - 12yz + 24xz
= (4x)² + (-2y)² + (3z)² + 2(4x)(-2y) + 2(-2y)(3z) + 2(3z)(4x)
= {4x + (-2y) + 3z}² [$\therefore$ a² + b² + c² + 2ab + 2bc + 2ca = (a + b + c)²]
= (4x - 2y + 3z)²
= (4x - 2y + 3z)(4x - 2y + 3z)

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Question 42 Marks

Without finding the cubes, factorise:
(x - 2y)³ + (2y - 3z)³ + (3z - x)³

Answer

We know that,
a³ + b³+ c³ - 3abc = (a + b + c) (a² + b² + c² - ab - bc - ca)
Also, if a + b + c = 0, then a³ + b³+ c³ = 3abc
Here, we see that (x - 2y) + (2y - 3z) + (3z - x) = 0
Therefore, (x - 2y)³ + (2y - 3z)³ + (3z - x)³ = 3(x - 2y)(2y - 3z)(3z - x) .

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Question 52 Marks

If $\text{p(x)}= \text{x}^2-4\text{x}+3,$ evaluate:
$\text{p}(2)-\text{p}(-1)+\text{p}\big(\frac{1}{2}\big).$

Answer

We have $\text{p(x)}= \text{x}^2-4\text{x}+3$
$\therefore$ $\text{p}(2)-\text{p}(-1)+\text{p}\Big(\frac{1}{2}\Big)$
$=(2^2-4\times2+3)-\big\{(-1)^2-4(-1)+3\big\}+\Big\{\Big(\frac{1}{2}\Big)^2-4\times\frac{1}{2}+3\Big\}$
$=(4-8+3)-(1+4+3)+\Big(\frac{1}{4}-2+3\Big)$
$=-1-8+\frac{5}{4}$
$=-9+\frac{5}{4}=\frac{-36+5}{4}=\frac{-31}{4}$

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Question 62 Marks

By actual division, find the quotient and the remainder when the first polynomial is divided by the second polynomial : x⁴ + 1; x - 1

Answer

By acute division, we have

quotient = x³ + x² + x + 1

remainder = 2

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Question 72 Marks

Using suitable identity, evaluate the following:

101 × 102

Answer

101 × 102 = (100 + 1)(100 + 2)

Now using identity (x + a)(x + b) = x² + (a + b)x + ab, we have

(100 + 1)(100 + 2) = (100)² + (1 + 2)100 + (1)(2)

= 10000 + (3)100 + 2 = 10000 + 300 + 2

= 10302

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Question 82 Marks

Expand the following:
(-x + 2y - 3z)²

Answer

(-x + 2y - 3z)² = {(-x) + 2y + (-3z)}²

= (-x)² + (2y)²+ (-3z)²+ 2(-x)(2y) + 2(2y)(-3z) + 2(-3yz)(-x)

= x² + 4y² + 9z² - 4xy - 12yz + 6xz.

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Question 92 Marks

Check whether p(x) is a multiple of g(x) or not:
p(x) = x³ - 5x² + 4x - 3, g(x) = x - 2

Answer

p(x) will be a multiple g(x) if g(x) divides p(x).
Now, g(x) = x - 2 gives x = 2
Remainder = p(2) = (2)³ - 5(2)² + 4(2) - 3
= 8 - 5(4) + 8 - 3 = 8 - 20 + 8 - 3
= -7
Since remainder ≠ 0, So p(x) is not a multiple of g(x).

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Question 102 Marks

Without actually calculating the cubes, find the value of:
$\Big(\frac{1}{2}\Big)^3+\Big(\frac{1}{3}\Big)^3-\Big(\frac{5}{6}\Big)^3$

Answer

Given, $\Big(\frac{1}{2}\Big)^3+\Big(\frac{1}{3}\Big)^3-\Big(\frac{5}{6}\Big)^3$ or $\Big(\frac{1}{2}\Big)^3+\Big(\frac{1}{3}\Big)^3+\Big(-\frac{5}{6}\Big)^3$

Here, we see that, $\frac{1}{2}+\frac{1}{3}-\frac{5}{6}=\frac{3+2-5}{6}=\frac{5-5}{6}=0$

$\therefore \ \Big(\frac{1}{2}\Big)^3+\Big(\frac{1}{3}\Big)^3-\Big(\frac{5}{6}\Big)^3=3\times\frac{1}{2}\times\frac{1}{3}\times\Big(-\frac{5}{6}\Big)=-\frac{5}{12}$

[Using identity, if a + b + c = 0, then a³ + b³+ c³ = 3abc]

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Question 112 Marks

Find the following product:
(x² - 1) (x⁴ + x² + 1)

Answer

(x² - 1)(x⁴ + x² + 1)

= (x² - 1) {(x²)² + (x²)(1) + (1)²}

= (x²)³ - (1)³

$\big[\therefore$ (a + b)(a² - ab + b²) = a³ + b³$\big]$

= x⁶ - 1

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Question 122 Marks

Expand the following:
(3a - 5b - c)²

Answer

(3a - 5b - c)² = (3a)² + (-5b)² + (-c)² + 2(3a)² - 5b + 2(-5b)(-c) + 2(-c)(3a)
[$\therefore$ a² + b² + c² + 2ab + 2bc + 2ca = (a + b + c)²]
= 9a² + 25b² + c² - 30ab + 10bc + 6ca.

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Question 132 Marks

Expand the following:
(4a - b + 2c)²

Answer

(4a - b + 2c)² = (4a)² + (-b)² + (2c)² + 2(4a)(-b) + 2(-b)(2c) + 2(2c)(4a)
[$\therefore$ a² + b² + c² + 2ab + 2bc + 2ca = (a + b + c)²]
= 16a² + b² + 4c² - 8ab - 4ac + 16ca

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Question 142 Marks

Find the value of:
x³ - 8y³ - 36xy - 216, when x = 2y + 6

Answer

Here, we see that, x - 2y - 6 = 0

$\therefore$ x³ + (-2y)³+ (-6)³ = 3x(-2y)(-6)

[Using identity, if a + b + c = 0, then a³ + b³+ c³ = 3abc]

⇒ x³ - 8y³ - 216 = 36xy ...(i)

Now, x³ - 8y³ - 36xy - 216

= x³ - 8y³ - 216 - 36xy

= 36xy - 36xy = 0 [From Eq...(i)]

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Question 152 Marks

By Remainder Theorem find the remainder, when p(x) is divided by g(x), where:
$\text{p(x)}=\text{x}^3-6\text{x}^2+2\text{x}-4,$ and $\text{g(x)}=1-\frac{3}{2}\text{x} $

Answer

Given, $\text{p(x)}=\text{x}^3-6\text{x}^2+2\text{x}-4,$ and $\text{g(x)}=1-\frac{3}{2}\text{x} $

Here, zero of g(x) is $\frac{2}{3}.$

When we divide p(x) by g(x) using remainder theorem, we get the remainder $\text{p}\big(\frac{2}{3}\big)$

$\therefore =\frac{8}{27}-6\times\frac{4}{9}+2\times\frac{2}{3}-4=\frac{8}{27}-\frac{24}{9}+\frac{4}{3}-4$

$=\frac{8-72+36-108}{27}=\frac{-136}{27}$

Hence, remainder is $\frac{-136}{27}.$

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Question 162 Marks

Factorise:
a³ - 8b³ - 64c³ - 24abc

Answer

We have,
a³ - 8b³ - 64c³ - 24abc
= {(a)³ + (-2b)³ + (-4c)³ - 3(a)(-2b)(-4c)}
= {a + (-2b) + (-4c)}{a² + (-2b)² + (-4c)² - a(-2b) - (-2b)(-4c) - (-4c)a}
$\big[\therefore$ a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)$\big]$
= (a + -2b + -4c)(a² + 4b² + 16c² + 2ab + 8bc + 4ca)

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Question 172 Marks

Factorise the following:
25x² + 16y² + 4z² - 40xy + 16yz - 20xz

Answer

25x² + 16y² + 4z² - 40xy + 16yz - 20xz

= (-5x)² + (4y)² + (2z)² + 2(-5x)(4y) + 2(4y)(2z) + 2(2z)(-5x)

= (-5x + 4y + 2z)²

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Question 182 Marks

Factorise the following:
$\Big(2\text{x}+\frac{1}{3}\Big)^2-\Big(\text{x}-\frac{1}{2}\Big)^2$

Answer

$\Big(2\text{x}+\frac{1}{3}\Big)^2-\Big(\text{x}-\frac{1}{2}\Big)^2$
$=\Bigg[\Big(2\text{x}+\frac{1}{3}\Big)-\Big(\text{x}-\frac{1}{2}\Big)\Bigg]\Bigg[\Big(2\text{x}+\frac{1}{3}\Big)+\Big(\text{x}-\frac{1}{2}\Big)\Bigg]$
[Using identity, a² - b² = (a - b)(a + b)]
$=\Big(2\text{x}-\text{x}+\frac{1}{3}+\frac{1}{2}\Big)\Big(2\text{x}+\text{x}+\frac{1}{3}-\frac{1}{2}\Big)$
$=\Big(\text{x}+\frac{2+3}{6}\Big)\Big(3\text{x}+\frac{2-3}{6}\Big)$
$=\Big(\text{x}+\frac{5}{6}\Big)\Big(3\text{x}-\frac{1}{6}\Big)$

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Question 192 Marks

Find the following product:
$\text{a}^3 - 2\sqrt{2}\text{b}^3$

Answer

$\text{a}^3 - 2\sqrt{2}\text{b}^3=(\text{a})^3 - (\sqrt{2}\text{b})^3$
$(\text{a} - \sqrt{2}\text{b})\big\{(\text{a})^2 +(\text{a}) (\sqrt{2}\text{b})+(\sqrt{2}\text{b})^2\big\}$
$\big[\therefore$ a³ - b³ = (a - b)(a² + ab + b²)$]$
$(\text{a} - \sqrt{2}\text{b})(\text{a}^2 + \sqrt{2}\text{ab}+2\text{b}^2)$

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Question 202 Marks

Factorise the following:
9x² + 4y² + 16z² + 12xy - 16yz - 24xz

Answer

9x² + 4y² + 16z² + 12xy - 16yz - 24xz
= (3x)² + (2y)² + (-4z)² + 2(3x)(2y) + 2(y)(-4z) + 2(-4z)(3x)
= {3x + 2y (-4z)}² [$\because$ a² + b² + c² + 2ab + 2bc + 2ca = (a + b + c)²]
= (3x + 2y - 4z)2 = (3x + 2y - 4z)(3x + 2y - 4z)

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Question 212 Marks

Factorise the following:
9x² - 12x + 4

Answer

9x² - 12x + 4 = (3x)² - 2(3x)(2) + (2)²
= (3x - 2)² [$\therefore$ a² - 2ab + b² = (a - b)²]
= (3x - 2)(3x - 2)

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Question 222 Marks

Factorise:
$2\sqrt{2}\text{a}^3 + 8\text{b}^3 - 27\text{c}^3 + 18\sqrt{2}\text{abc}.$

Answer

We have,
$2\sqrt{2}\text{a}^3 + 8\text{b}^3 - 27\text{c}^3 + 18\sqrt{2}\text{abc}.$
$=\big\{(\sqrt{2}\text{a})^3 + (2\text{b})^3 + (-3\text{c})^3 - 3(\sqrt{2}\text{a})(2\text{b})(-3\text{c})\big\}$
$=\big\{\sqrt{2}\text{a} + 2\text{b} -3\text{c}\big\}\big\{(\sqrt{2}\text{a})^2+(2\text{b})^2+(-3\text{c})^2-(\sqrt{2}\text{a})(2\text{b})-(2\text{b})(-3\text{c})(\sqrt{2}\text{a})\big\}$
$=(\sqrt{2}\text{a} + 2\text{b} -3\text{c})(2\text{a}^2+4\text{b}^2+9\text{c}^2-2\sqrt{2}\text{ab}+6\text{bc}+3\sqrt{2}\text{ca})$

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Question 232 Marks

For what value of m is x³ - 2mx² + 16 divisible by x + 2 ?

Answer

Let p(x) = x³ - 2mx² + 16
Since, p(x) is divisible by (x + 2), then remainder = 0
P(-2) = 0
⇒ (-2)³ - 2m(-2)² + 16 = 0
⇒ -8 - 8m + 16 = 0
⇒ 8 = 8m
m = 1
Hence, the value of m is 1 .

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Question 242 Marks

Give possible expressions for the length and breadth of the rectangle whose area is given by 4a² + 4a - 3.

Answer

Given, area of rectangle = 4a² + 6a - 2a - 3
= 4a² + 4a - 3 [by splitting middle term]
= 2a(2a + 3) - 1(2a + 3) = (2a - 1)(2a + 3)
Hence, possible length = 2a - 1 and breadth = 2a + 3

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Question 252 Marks

Find the following product:
$\Big(\frac{\text{x}}{2}+2\text{y}\Big)\Big(\frac{\text{x}^2}{4}-\text{xy}+4\text{y}^2\Big)$

Answer

$\Big(\frac{\text{x}}{2}+2\text{y}\Big)\Big(\frac{\text{x}^2}{4}-\text{xy}+4\text{y}^2\Big)$
$=\Big(\frac{\text{x}}{2}+2\text{y}\Big)\Big\{\Big(\frac{\text{x}}{2}\Big)^2-\Big(\frac{\text{x}}{2}\Big)(\text{2y})+(2\text{y})^2\Big\}$
$=\Big(\frac{\text{x}}{2}\Big)^3+(2\text{y})^3$ $\big[\therefore$ (a + b)(a² - ab + b²) = a³ + b³$\big]$
$=\frac{\text{x}}{8}^3+8\text{y}^3$

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Question 262 Marks

Show that:
x + 3 is a factor of 69 + 11x - x² + x³.

Answer

Let p(x) = 69 + 11x - x² + x³, g(x) = x + 3.
g(x) = x + 3 = 0 gives x = -3
g(x) will be a factor of p(x) if p(-3) = 0(Factor theorem)
Now, p(-3) = 69 + 11(-3) - (-3)² + (-3)³
= 69 - 33 - 9 -27
= 0
Since, p(-3) = 0, So g(x) is a factor of p(x).

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Question 272 Marks

Factorise:
1 + 64x³

Answer

We have,
1 + 64x³ = (1)³ + (4x)³
= (1 + 4x){(1)² - (1)(4x) + (4x)²}
$\big[\therefore$ a³ + b³ = (a + b)(a² - ab + b²)$\big]$
= (1 + 4x)(1 - 4x + 16x²)
= (1 + 4x)(16x² -4x + 1)
= (4x + 1)(16x² -4x + 1)

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Question 282 Marks

Find the zeroes of the polynomial:
p(x) = (x - 2)² - (x + 2)²

Answer

Given, polynomial is p(x) = (x - 2)² - (x + 2)²
For zeroes of polynomial, put p(x) = 0
(x – 2)² – (x+ 2)² = 0
(x - 2 + x + 2)(x - 2 - x - 2) = 0 [using identity, a²- b² = (a - b)(a + b)] ⇒ (2x)(-4) = 0

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Question 292 Marks

Factorise the following:
9x² - 12x + 3

Answer

9x² - 12x + 3 = 9x² - 9x - 3x + 3
= 9x(x - 1) - 3(x - 1)
= (9x - 3)(x - 1)
= 3(3x - 1)(x - 1)

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Question 302 Marks

By Remainder Theorem find the remainder, when p(x) is divided by g(x), where:
p(x) = 4x³ - 12x² + 14x - 3, g(x) = 2x - 1

Answer

Given, p(x) = 4x³ - 12x² + 14x - 3, g(x) = 2x - 1
Here, zero of g(x) is $\frac{1}{2}.$
When we divide p(x) by g(x) using remainder theorem, we get the remainder $\text{p}\big(\frac{1}{2}\big)$
$\therefore\text{p}\big(\frac{1}{2}\big)=4\big(\frac{1}{2}\big)^3-12\big(\frac{1}{2}\big)^2+14\big(\frac{1}{2}\big)-3$
$=4\times\frac{1}{8}-12\times\frac{1}{4}+14\times\frac{1}{2}-3$
Hence, remainder is $\frac{3}{2}.$

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Question 312 Marks

Find the value of:
x³ + y³ - 12xy + 64, when x + y = -4

Answer

Here, we see that, x + y + 4 = 0
$\therefore$ x³ + y³+ (4)³ = 3xy(4)
[Using identity, if a + b + c = 0, then a³ + b³+ c³ = 3abc]
= 12xy ...(i)
Now, x³ + y³ - 12xy + 64 = x³ + y³+ 64 - 12xy
= 12xy - 12xy = 0 [From Eq...(i)]

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Question 322 Marks

By Remainder Theorem find the remainder, when p(x) is divided by g(x), where:
p(x) = x³ - 2x² - 4x - 1, g(x) = x + 1

Answer

Given, p(x) = x³ - 2x² - 4x - 1 and g(x) = x + 1
Here, zero of g(x) is -1.
When we divide p(x) by g(x) using remainder theorem, we get the remainder p(-1)
$\therefore$ p(-1) = (-1)³ - 2(-1)² - 4(-1) - 1
= -1 - 2 + 4 - 1
= 4 - 4 = 0
Hence, remainder is 0.

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Question 332 Marks

Without actually calculating the cubes, find the value of:
(0.2)³ - (0.3)³ + (0.1)³

Answer

Given, (0.2)³ - (0.3)³ + (0.1)³ or (0.2)³ + (-0.3)³ + (0.1)³
Here, we see that, 0.2 - 0.3 + 0.1 = 0.3 - 0.3 = 0
$\therefore$ (0.2)³ - (0.3)³ + (0.1)³ = 3 × (0.2) × (-0.3) × (0.1)
[Using identity, if a + b + c = 0, then a³ + b³+ c³ = 3abc]
= -0.6 × 0.003 = -0.018

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Question 342 Marks

By Remainder Theorem find the remainder, when p(x) is divided by g(x), where:
p(x) = x³ - 3x² + 4x + 50, g(x) = x - 3

Answer

Given, p(x) = x³ - 3x² + 4x + 50, g(x) = x - 3
Here, zero of g(x) is 3.
When we divide p(x) by g(x) using remainder theorem, we get the remainder p(3)
$\therefore$ p(3) = (3)³ - 3(3)² + 4(3) + 50
= 27 - 27 + 12 + 50 = 62
Hence, remainder is 62.

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Question 352 Marks

Using suitable identity, evaluate the following:
103³

Answer

103³ = (100 + 3)³
Now using identity (a + b)³ = a³ + b³ + 3ab(a + b), we have
(100 + 3)³ = (100)³ + (3)³ + 3(100)(3)(100 + 3)
= 1000000 + 27 + 900(100 + 3)
= 1000000 + 27 + 90000 + 2700
= 1092727

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Question 362 Marks

Expand the following:
$\Big(\frac{1}{\text{x}}+\frac{\text{y}}{3}\Big)^3$

Answer

$\Big(\frac{1}{\text{x}}+\frac{\text{y}}{3}\Big)^3=\Big(\frac{1}{\text{x}}\Big)^3 + \Big(\frac{\text{y}}{3}\Big)^3 + 3\Big(\frac{1}{\text{x}}\Big)\Big(\frac{\text{y}}{3}\Big)\Big(\frac{1}{\text{x}}+\frac{\text{y}}{3}\Big)$
[Using identity, (a - b)³ = a³ - b³ + 3a(-b)(a - b)]
$=\frac{1}{\text{x}^3}+\frac{\text{y}^3}{27}+\frac{\text{y}}{\text{x}}=\Big(\frac{1}{\text{x}}+\frac{\text{y}}{3}\Big)$
$=\frac{1}{\text{x}^3}+\frac{\text{y}^3}{27}+\frac{\text{y}}{\text{x}^2}+\frac{\text{y}^2}{3\text{x}}$

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Question 372 Marks

If a, b, c are all non-zero and a + b + c = 0, prove that $\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}=3.$

Answer

We have a, b, c are all non-zero and a + b + c = 0, therefore
a³ + b³+ c³ = 3abc
Now, $\frac{\text{a}^2}{\text{bc}}+\frac{\text{b}^2}{\text{ca}}+\frac{\text{c}^2}{\text{ab}}=\frac{\text{a}^3+\text{b}^3+\text{c}^3}{\text{abc}}=\frac{3\text{abc}}{\text{abc}}=3$

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Question 382 Marks

Expand the following:
$\Big(4-\frac{1}{3{\text{x}}}\Big)^3$

Answer

$\Big(4-\frac{1}{3{\text{x}}}\Big)^3=(4)^3+\Big(-\frac{1}{3\text{x}}\Big)^3+ 3(4)\Big(-\frac{1}{3\text{x}}\Big)\Big(4-\frac{1}{3{\text{x}}}\Big)$
[Using identity, (a - b)³ = a³ - b³ + 3a(-b)(a - b)]
$= 64 - \frac{1}{27\text{x}^3}-\frac{4}{\text{x}}\Big(4-\frac{1}{3\text{x}}\Big)$
$= 64 - \frac{1}{27\text{x}^3}-\frac{16}{\text{x}}+\frac{4}{3\text{x}^2}$

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